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Proof step 5 from proof divides-abs-02 (previous step)

⊦ (∃ i0. i * i0 = abs(j) * j0) ↔ (∃ i0. i * i0 = j * j0)

User applied rule case distinction

And obtained the following new goals

Proof step 6:
∃ i0. i * i0 = abs(j) * j0 ⊦ ∃ i0. i * i0 = j * j0
Proof step 13:
∃ i0. i * i0 = j * j0 ⊦ ∃ i0. i * i0 = abs(j) * j0

Rule argument

pos: right 1

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